Question

Cho x,y,z,a,b,c>0

\(\sqrt[3]{abc}+\sqrt[3]{xyz}\le\sqrt[3]{(a+x)(b+y)(c+z)}\)

Answer 1

\(BDT\Leftrightarrow\sqrt[3]{\dfrac{abc}{\left(a+x\right)\left(b+y\right)\left(c+z\right)}}+\sqrt[3]{\dfrac{xyz}{\left(a+x\right)\left(b+y\right)\left(c+z\right)}}\le1\)

Áp dụng BĐT AM-GM ta có:

\(\sqrt[3]{\dfrac{abc}{(a+x)(b+y)(c+z)}}\le\dfrac{\dfrac{a}{a+x}+\dfrac{b}{b+y}+\dfrac{c}{c+z}}{3}\)

\(\sqrt[3]{\dfrac{xyz}{(a+x)(b+y)(c+z)}}\le\dfrac{\dfrac{x}{a+x}+\dfrac{y}{b+y}+\dfrac{z}{c+z}}{3}\)

Cộng theo vế 2 BĐT trên:

\(\Rightarrow VT\le\dfrac{\dfrac{x+a}{x+a}+\dfrac{b+y}{b+y}+\dfrac{c+z}{c+z}}{3}=1=VP\) *ĐPCM*

Answer 2

@Ace Legona